Question

求解一道不定积分，谢谢

∫（x^3+1）/（x^2+1）^2 dx

Answer 1

先计算∫ dx/(x² + 1)²，令x = tanz，dx = sec²z dz ==> √(x² + 1) = secz
= ∫ sec²z/sec⁴z dz
= ∫ cos²z dz
= ∫ (1 + cos2z)/2 dz
= (1/2)z + (1/2)sinzcosz + C
= (1/2)arctanz + (1/2)[x/√(x² + 1)][1/√(x² + 1)] + C
= (1/2)[arctanx + x/(x² + 1)] + C

∫ (x³ + 1)/(x² + 1)² dx
= ∫ [x(x² + 1 - 1) + 1]/(x² + 1)² dx
= ∫ x/(x² + 1) dx + ∫ (1 - x)/(x² + 1)² dx
= (1/2)∫ 1/(x² + 1) d(x² + 1) + ∫ dx/(x² + 1)² - (1/2)∫ 1/(x² + 1)² d(x² + 1)
= (1/2)ln(x² + 1) + 1/[2(x² + 1)] + ∫ dx/(x² + 1)²，代入上面的结果
= (1/2)ln(x² + 1) + 1/[2(x² + 1)] + (1/2)[arctanx + x/(x² + 1)] + C
= (1/2)[(x + 1)/(x² + 1) + ln(x² + 1) + arctanx] + C